Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
] 广度优先搜索，一般做法都是采取一个队列来做。
不过这题有个比较不一样的是他返回的格式是List<List>>的格式，所以需要做一些处理。
直接上代码了，很简单的题，
JAVA CODE:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder( TreeNode root ) { List<List<Integer>> returnList = new ArrayList<List<Integer>>(); if(root == null){
return returnList;
} List<Integer> temp = null; Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.add( root ); while( !queue.isEmpty() ) {
Queue<TreeNode> tempQ = new LinkedList<TreeNode>();
temp = new ArrayList<Integer>();
while( !queue.isEmpty() ) {
TreeNode tn = queue.poll(); if( tn.left != null ) {
tempQ.add( tn.left );
}
if( tn.right != null ) {
tempQ.add( tn.right );
}
temp.add( tn.val );
} queue = tempQ;
returnList.add( temp );
} return returnList;
}
}